POJ 3159 Candies（差分约束，最短路）

Candies

Time Limit: 1500MS		Memory Limit: 131072K
Total Submissions: 20067		Accepted: 5293

Description

During the kindergarten days, flymouse was the monitor of his class. Occasionally the head-teacher brought the kids of flymouse’s class a large bag of candies and had flymouse distribute them. All the kids loved candies very much and often compared the numbers of candies they got with others. A kid A could had the idea that though it might be the case that another kid B was better than him in some aspect and therefore had a reason for deserving more candies than he did, he should never get a certain number of candies fewer than B did no matter how many candies he actually got, otherwise he would feel dissatisfied and go to the head-teacher to complain about flymouse’s biased distribution.

snoopy shared class with flymouse at that time. flymouse always compared the number of his candies with that of snoopy’s. He wanted to make the difference between the numbers as large as possible while keeping every kid satisfied. Now he had just got another bag of candies from the head-teacher, what was the largest difference he could make out of it?

Input

The input contains a single test cases. The test cases starts with a line with two integers N and M not exceeding 30 000 and 150 000 respectively. N is the number of kids in the class and the kids were numbered 1 through N. snoopy and flymouse were always numbered 1 and N. Then follow M lines each holding three integers A, B and c in order, meaning that kid A believed that kid B should never get over c candies more than he did.

Output

Output one line with only the largest difference desired. The difference is guaranteed to be finite.

Sample Input

2 2
1 2 5
2 1 4

Sample Output

5

Hint

32-bit signed integer type is capable of doing all arithmetic.

Source

POJ Monthly--2006.12.31, Sempr

 

 

 

 

差分约束。

 

总共n个点，m对 A B c  表示B-A<=c

要求1和n的差值最大

每个约束B-A<=c 就是B<=A+c  加边A->B  为c的边。

 

建图以后就是求最短路。

 

优先队列+Dijkstra。原来用vector的模板超时，改数组就好了

 

//============================================================================
// Name        : POJ.cpp
// Author      :
// Version     :
// Copyright   : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <algorithm>
#include <queue>
#include <vector>
using namespace std;
/*
 * 使用优先队列优化Dijkstra算法
 * 复杂度O(ElogE)
 * 注意对vector<Edge>E[MAXN]进行初始化后加边
 */
const int INF=0x3f3f3f3f;
const int MAXN=;
struct qnode
{
    int v;
    int c;
    qnode(int _v=,int _c=):v(_v),c(_c){}
    bool operator <(const qnode &r)const
    {
        return c>r.c;
    }
};
struct Edge
{
    int v,cost;
    int next;
};
Edge edge[];
int tol;
int head[MAXN];
bool vis[MAXN];
int dist[MAXN];
void Dijkstra(int n,int start)//点的编号从1开始
{
    memset(vis,false,sizeof(vis));
    for(int i=;i<=n;i++)dist[i]=INF;
    priority_queue<qnode>que;
    while(!que.empty())que.pop();
    dist[start]=;
    que.push(qnode(start,));
    qnode tmp;
    while(!que.empty())
    {
        tmp=que.top();
        que.pop();
        int u=tmp.v;
        if(vis[u])continue;
        vis[u]=true;
        for(int i=head[u];i!=-;i=edge[i].next)
        {
            int v=edge[i].v;
            int cost=edge[i].cost;
            if(!vis[v]&&dist[v]>dist[u]+cost)
            {
                dist[v]=dist[u]+cost;
                que.push(qnode(v,dist[v]));
            }
        }
    }
}
void addedge(int u,int v,int w)
{
    edge[tol].v=v;
    edge[tol].cost=w;
    edge[tol].next=head[u];
    head[u]=tol++;
}

int main()
{
//    freopen("in.txt","r",stdin);
//    freopen("out.txt","w",stdout);
    int n,m;
    while(scanf("%d%d",&n,&m)==)
    {
        tol=;
        memset(head,-,sizeof(head));
        int A,B,C;
        while(m--)
        {
            scanf("%d%d%d",&A,&B,&C);
            addedge(A,B,C);
        }
        Dijkstra(n,);
        printf("%d\n",dist[n]);
    }
    return ;
}